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JEE Advanced 2014

JEE Main 2014 Mathematics question on Definite Integrals and Areas

The integral ∫π/4π/2(2cosec⁡x)17dx\int_{\pi/4}^{\pi/2} (2\cosec x)^{17} dx is equal to
  1. A.
    ∫0log⁡(1+2)2(eu+e−u)16du\int_0^{\log(1+\sqrt{2})} 2(e^u + e^{-u})^{16} du
  2. B.
    ∫0log⁡(1+2)(eu+e−u)17du\int_0^{\log(1+\sqrt{2})} (e^u + e^{-u})^{17} du
  3. C.
    ∫0log⁡(1+2)(eu−e−u)17du\int_0^{\log(1+\sqrt{2})} (e^u - e^{-u})^{17} du
  4. D.
    ∫0log⁡(1+2)2(eu−e−u)16du\int_0^{\log(1+\sqrt{2})} 2(e^u - e^{-u})^{16} du
Show answer and solution

Correct option: A

Solution

Substitute u = ln(csc x - cot x), so that eu = csc x - cot x and e^(-u) = csc x + cot x (since (csc x - cot x)(csc x + cot x) = 1). Then eu + e^(-u) = 2 csc x, and du = (csc x) dx (since d/dx[ln(csc x - cot x)] = csc x).

When x = π/4: csc x - cot x = √2 - 1, so u = ln(√2 - 1) = -ln(1 + √2). When x = π/2: csc x - cot x = 1 - 0 = 1, so u = 0. Flipping limits absorbs the minus sign.

The integral becomes ∫ (2 csc x)17 dx = ∫ (eu + e^(-u))17 · (1/csc x) du = ∫ (eu + e^(-u))17 · (sin x) du. Since 2 csc x = eu + e^(-u), sin x = 2/(eu + e^(-u)), so the integrand is (eu + e^(-u))17 · 2/(eu + e^(-u)) = 2(eu + e^(-u))16.

The result is ∫₀^(ln(1+√2)) 2(eu + e^(-u))16 du, which is option 1.

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